Football Release Height: 19.0m/s, 30.5deg, 35.0m Away

[brunettegurl]

Homework Statement

A quarterback throws a football with an initial speed of 19.0 m/s at an angle 30.5deg above the horizontal. Unfortunately, the pass is incomplete and the football drops to the ground, 35.0 m away from the quaterback. From what height was the football released? (no diagram provided)

Homework Equations

Vy= Vknoty - gt from which we get t_tot= [tex]\frac{2V_osin\vartheta}{g}[/tex]
from the equation y= V_oyt_y2-0.5gt_0.5² we get hmax= v₀²sin²g$$\vartheta$$/2g

The Attempt at a Solution

so to figure out far it goes i used the equation xmax= v_0xt_tot where it wld now equal vknotcos$$\vartheta$$(2*vknot sin$$\vartheta$$/g) + distance it traveled and i wanted to know what i was doing wrong thanks

 

[Cyosis]

Not sure what you're doing here. You set v_y=0 when the ball touches the ground? This is not true for if there had been no ground it would have continued to fall.

You know the ball traveled a distance of 35 meters in the x-direction. So you can calculate how long it took for the ball to reach that position with $x=v_x t$. Knowing the flight time you can solve the initial height by using $y=y_0+v_0 t-1/2 gt^2$.

 

[brunettegurl]

why wld the velocity be v₀ and not V_y??

 

[Cyosis]

It's not I just listed the general kinematic expression for a falling object. In your case the vertical starting velocity is v_y.

 

[brunettegurl]

thanx